I use Locust, a load testing framework, and the following is the summary of a test result.
Name # reqs # fails Avg Min Max | Median req/s
--------------------------------------------------------------------------------------------------------------------------------------------
GET /sample 10000 0(0.00%) 97 56 349 | 96 761.90
--------------------------------------------------------------------------------------------------------------------------------------------
Total 10000 0(0.00%) 761.90
I guess that req/s means 761.90 requests are processed in 1 second. How about Avg, Min, Max and Median? How can I read these columns?.
This perfomance test tooks about 15 sec. I set min_wait = 0 max_wait = 0.
Looking at the source, it appears to refer to the response time
Related
I'm kinda new to DAX and PowerBi and I need to translate my SQL IF statement for whatever syntax is this on PowerBi to achieve the output I want.
Sql code I want to translate:
IF (Payment.payment>0) AND (Account.PV = Account.GV) THEN 1 ELSE 0
I want to make a calculated column on Payment table which will return 1 or 0 so that I can use this to filter all the records that meets my condition
account_id is the relationship of these two tables
Here is a sample data for reference: Account table
account_id
pv
gv
due_date
123
100
200
08/08/2022
124
200
200
08/09/2022
125
300
800
08/10/2022
126
400
670
08/11/2022
127
500
500
08/12/2022
128
600
600
08/13/2022
129
700
1000
08/14/2022
130
800
760
08/15/2022
131
900
900
08/16/2022
132
1000
1000
08/17/2022
133
1100
2300
08/09/2022
Here is a sample data for reference: Payment table
payment_id
payment_number
payment
payment_date
account_id
_test
101
554321
1000
03/01/2022
123
0
102
554322
1200
03/21/2022
124
1
103
554322
1100
04/28/2022
124
1
104
554323
2500
05/04/2022
131
1
105
554324
3000
05/14/2022
133
0
106
554325
3000
05/14/2022
132
1
107
554322
1200
03/21/2022
124
1
108
554323
2500
04/05/2022
131
1
109
554328
3100
04/05/2022
128
0
Codes I tried but I can't help myself to find the correct way to do it correctly and return the output that I need
_test = IF(Payments[payment]>0 && RELATED('Account'[PV])=RELATED('Account'[GV]), 1)
_test = IF(AND(Payments[payment])>0, RELATED('Account'[PV])=RELATED('Account'[GV])),1,0)
Any suggestion is much appreciated. Please recommend what kind of syntax/function should be used in order to achieve the output or what would be the work around to use other than IF statement
The problem that you are facing with RELATED is that RELATED only works from 1 side to many side.
Meaning, that if you bring the axis from 1-side and perform a calculation on the many side the filter works perfectly. Take a look at the direction of the filter below. The direction of the filter tells you on normal circumstances, you should bring your axis from Account and whatever calculation you perform on `Payment table it will work out.
But you are doing exactly the reverse. You are bringing the axis from Payment and hoping for RELATED to work. It won't cause the direction to be as such.
However, DAX is much more dynamic than that. If for whatever reason, you need to bring axis from many side where you need to still filter on 1-side, you can define a reverse filter direction on-the-fly (because DAX is magical) without needing to change anything in the data model by using CROSSFILTER. With CROSSFILTER you are customizing the filter direction as such
CROSSFILTER(<LEFtblcolumnName1>, <RIGHTtblcolumnName2>, <direction>)
This is how, (with your given dataset)
Column =
VAR cond1 =
CALCULATE (
MAX ( Account[Account.pv] ),
CROSSFILTER ( Payment[Payment.account_id], Account[Account.account_id], BOTH )
)
- CALCULATE (
MAX ( Account[Account.gv] ),
CROSSFILTER ( Payment[Payment.account_id], Account[Account.account_id], BOTH )
)
RETURN
IF ( cond1 == 0 && Payment[Payment.payment] > 0, 1, 0 )
I have a very large dataset with 'user_id', 'event_id' and 'event_date'.
What would be the most efficient way in scala spark, to calculate the events' co-occurence matrix, with a co-occurence being counted only if happening within a window of say n days for a particular customer ?
user_id
event_id
date
1
100
01/01/2000
1
101
05/01/2000
1
102
14/05/2021
1
103
20/05/2021
2
100
04/06/2001
2
101
10/06/2001
2
102
13/08/2021
2
103
30/08/2021
So, if we consider a 15 days window, for each user we obtain the following co-occurences:
user_id
event_id_1
event_id_2
1
100
101
1
102
103
2
100
101
Hence, finally the total number of co-occurences:
event_id_1
event_id_2
count
100
101
2
102
103
1
That's the process I need to run in Scala Spark. Any suggestion?
Many thanks in advance
Initial Table
company time value
-------------------------
a 00:00:15.000 100
a 00:00:30.000 100
b 00:01:00.000 100
a 00:01:10.000 100
a 00:01:15.000 100
a 00:01:20.000 300
a 00:01:25.000 100
b 00:01:30.000 400
a 00:01:50.000 100
a 00:02:00.000 100
a 00:00:03.000 200
Let t = 1 hour.
For each row, I would like to look back t time.
Entries falling in t will form a time window. I would like to get max(time window) - min (time window) / number of events).
For example, if it is 12:00 now, and there are a total of five events, 12:00, 11:50, 11:40, 11:30, 10:30, four of which falls in the window of t i.e. 12:00, 11:50, 11:40, 11:30, the result will be 12:00 - 11:30 / 4.
Additionally, the window should only account for rows with the same value and company name.
Resultant Table
company time value x
--------------------------------
a 00:00:15.000 100 0 (First event A).
a 00:00:30.000 100 15 (30 - 15 / 2 events).
b 00:01:00.000 100 0 (First event of company B).
a 00:01:10.000 100 55/3 = 18.33 (1:10 - 0:15 / 3 events).
a 00:01:15.000 100 60/4 = 15 (1:15 - 0:15 / 4 events).
a 00:01:20.000 300 0 (Different value).
a 00:01:25.000 100 55/4 = 13.75 (01:25 - 0:30 / 4 events).
b 00:01:30.000 400 0 (Different value and company).
a 00:01:50.000 100 40/4 = 10 (01:50 - 01:10 / 4 events).
a 00:02:00.000 100 50/5 = 10 (02:00 - 01:10 / 5 events).
a 00:03:00.000 200 0 (Different value).
Any help will be greatly appreciated. If it helps, I asked a similar question, which worked splendidly: Sum values from the previous N number of days in KDB?
Table Query
([] company:`a`a`b`a`a`a`a`b`a`a`a; time: 00:00:15.000 00:00:30.000 00:01:00.000 00:01:10.000 00:01:15.000 00:01:20.000 00:01:25.000 00:01:30.000 00:01:50.000 00:02:00.000 00:03:00.000; v: 100 100 100 100 100 300 100 400 100 100 200)
You may wish to use the following;
q)update x:((time-time[time binr time-01:00:00])%60000)%count each v where each time within/:flip(time-01:00:00;time) by company,v from t
company time v x
---------------------------------
a 00:15:00.000 100 0
a 00:30:00.000 100 7.5
b 01:00:00.000 100 0
a 01:10:00.000 100 18.33333
a 01:15:00.000 100 15
a 01:20:00.000 300 0
a 01:25:00.000 100 13.75
b 01:30:00.000 400 0
a 01:50:00.000 100 10
a 02:00:00.000 100 10
a 03:00:00.000 200 0
It uses time binr time-01:00:00 to get the index of the min time for the previous 1 hour of each time.
Then (time-time[time binr time-01:00:00])%60000 gives the respective time range (i.e., time - min time) for each time in minutes.
count each v where each time within/:flip(time-01:00:00;time) gives the number of rows within this range.
Dividing the two and implementing by company,v applies it all only to those that have the same company and v values.
Hope this helps.
Kevin
If your table is ordered by time then below solution will give you the required result. You can also order your table by time if it is not already using xasc.
I have also modified the table to have time with different hour values.
q) t:([] company:`a`a`b`a`a`a`a`b`a`a`a; time: 00:15:00.000 00:30:00.000 01:00:00.000 01:10:00.000 01:15:00.000 01:20:00.000 01:25:00.000 01:30:00.000 01:50:00.000 02:00:00.000 03:00:00.000; v: 100 100 100 100 100 300 100 400 100 100 200)
q) f:{(`int$x-x i) % 60000*1+til[count x]-i:x binr x-01:00:00}
q) update res:f time by company,v from t
Output
company time v res
---------------------------------
a 00:15:00.000 100 0
a 00:30:00.000 100 7.5
b 01:00:00.000 100 0
a 01:10:00.000 100 18.33333
a 01:15:00.000 100 15
a 01:20:00.000 300 0
a 01:25:00.000 100 13.75
b 01:30:00.000 400 0
a 01:50:00.000 100 10
a 02:00:00.000 100 10
a 03:00:00.000 200 0
You can modify the function f to change time window value. Or change f to accept that as an input parameter.
Explanation:
We pass time vector by company, value to a function f. It deducts 1 hour from each time value and then uses binr to get the index of the first time entry within 1-hour window range from the input time vector.
q) i:x binr x-01:00:00
q) 0 0 0 0 1 2 2
After that, it uses the indexes of the output to calculate the total count. Here I am multiplying the count by 60000 as time differences are in milliseconds because it is casting it to int.
q) 60000*1+til[count x]-i
q) 60000 120000 180000 240000 240000 240000 300000
Then finally we subtract the min and max time for each value and divide them by the above counts. Since time vector is ordered(ascending), the input time vector can be used as the max value and min values are at indexes referred by i.
q) (`int$x-x i) % 60000*1+til[count x]-i
Hi lets say that i have matrix size 5x5.
B=[1 2 3 4 5; 10 20 30 40 50; 100 200 300 400 500; 1000 2000 3000 4000 5000; 10000 20000 30000 40000 50000];
How do i use function sum, to sum rows between 2 and 4 and have result:
A = [1110;2220;3330;4440]
You'll find some useful information about matrix indexing in the documentation at http://www.mathworks.co.uk/help/matlab/math/matrix-indexing.html
To illustrate your example, you can use B(2:4,:) to retreive the following:
ans =
10 20 30 40 50
100 200 300 400 500
1000 2000 3000 4000 5000
You can then use the sum function as follows to achieve your desired result:
A = sum(B(2:4,:))
I hope this helps!
All the best,
Matt
MATLAB>> sum(B(2:4,1:4))
ans =
1110 2220 3330 4440
If you want to transpose the result, add ' at the end.
I am doing some simulations studies and for initial stuides I am trying to simulate 100 gas particles and then grouping of these gas particles in 5 groups randomly for 10 or 100 times (non zero values in any groups). after that i have to find the group with highest particle and the number.
for example
100 gas particles
1 2 3 4 5(groups) Total particle group/Highest number
20|20|20|20|20 100 1-2-3-4-5/20
70|16|04|01|09 100 1/70
18|28|29|10|15 100 3/29
.
.
etc
i have used this to generate 5 random numbers for a single time
for i=1:1
randi([1,100],1,5)
end
ans =
50 41 9 60 88
but how will i find the highest number and group?
Use the max function :
a = [50 41 9 60 88];
[C,I] = max(a)
C should be equal to 88 and I to 4.
For the special case of equality (first line in your code), you have to read the documentation to see the result of max. I think the index returned will be the first max.